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Chapter 13: Problem 53

At a particular temperature, 12.0 moles of \(\mathrm{SO}_{3}\) is placed into a 3.0 -L rigid container, and the \(\mathrm{SO}_{3}\) dissociates by the reaction $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$At equilibrium, 3.0 moles of \(\mathrm{SO}_{2}\) is present. Calculate \(K\) for this reaction.

Short Answer

Expert verified
The equilibrium constant $K$ for the reaction $2 \mathrm{SO_{3}(g)} \rightleftharpoons 2 \mathrm{SO_{2}(g)} + \mathrm{O_{2}(g)}$ under the given conditions is 4.5.

Step by step solution

01

Write the reaction and determine the initial concentrations

The given reaction is: \[2 \mathrm{SO_{3}(g)} \rightleftharpoons 2 \mathrm{SO_{2}(g)} + \mathrm{O_{2}(g)}\] We are given initial moles of \(\mathrm{SO_{3}}\) and volume, therefore, we can calculate the initial concentration of \(\mathrm{SO_{3}}\) using the formula: Concentration = Moles/Volume \[[\mathrm{SO_{3}}]_0 = \frac{12.0\: \text{moles}}{3.0\: \text{L}} = 4.0\: M\] As no other amounts have been specified, we assume that initially there are no \(\mathrm{SO_{2}}\) and \(\mathrm{O_{2}}\).
02

Write an expression for the changes in concentrations

Assume that x moles of \(\mathrm{SO_{3}}\) dissociates. Since the stoichiometry is 2:2:1, we see that the change in concentration of \(\mathrm{SO_{2}}\) will be 2x and that of \(\mathrm{O_{2}}\) will be x. So the changes in concentrations for the reaction are: \[\begin{cases} \Delta [\mathrm{SO_{3}}] = -2x \\ \Delta [\mathrm{SO_{2}}] = 2x \\ \Delta [\mathrm{O_{2}}] = x \end{cases}\]
03

Form the equilibrium expressions

At equilibrium, we can write the concentrations as follows: \[[\mathrm{SO_{3}}]_{eq} = 4.0\: M - 2x\] \[[\mathrm{SO_{2}}]_{eq} = 2x\] \[[\mathrm{O_{2}}]_{eq} = x\] We are given that there are 3.0 moles of \(\mathrm{SO_{2}}\) at equilibrium. We will use this information to find the value of x. \[2x = 3.0\: \text{moles}\] \[x = 1.5\: \text{moles}\] Now, we can find the equilibrium concentrations by plugging in the value of x to the equilibrium expressions.
04

Calculate the equilibrium concentrations

Using the value of x we just found, we can calculate the equilibrium concentrations: \[[\mathrm{SO_{3}}]_{eq} = 4.0\: M - 2(1.5\: \text{moles}) = 1.0\: M\] \[[\mathrm{SO_{2}}]_{eq} = 2(1.5\: \text{moles}) = 3.0\: M\] \[[\mathrm{O_{2}}]_{eq} = 1.5\: \text{moles} = 0.5\: M\]
05

Determine the equilibrium constant (K)

We can now determine the equilibrium constant K for the reaction by forming the expression for K and plugging in the equilibrium concentrations we found in the previous step: \[K = \frac{[\mathrm{SO_{2}}]^2 \times [\mathrm{O_{2}}]}{[\mathrm{SO_{3}}]^2}\] Substituting the equilibrium concentrations: \[K = \frac{(3.0\: M)^2 \times (0.5\: M)}{(1.0\: M)^2}\] \[K = \frac{9.0 \times 0.5}{1.0}\] \[K = 4.5\] So, the equilibrium constant K for this reaction is 4.5.

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Most popular questions from this chapter

The value of the equilibrium constant \(K\) depends on which of the following (more than one answer may be correct)? a. the initial concentrations of the reactants b. the initial concentrations of the products c. the temperature of the system d. the nature of the reactants and products Explain.

At high temperatures, elemental nitrogen and oxygen react with each other to form nitrogen monoxide: $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$ Suppose the system is analyzed at a particular temperature, and the equilibrium concentrations are found to be \(\left[\mathrm{N}_{2}\right]=\) \(0.041 M,\left[\mathrm{O}_{2}\right]=0.0078 M,\) and $[\mathrm{NO}]=4.7 \times 10^{-4} M .\( Calculate the value of \)K$ for the reaction.

Consider an equilibrium mixture of four chemicals (A, B, C, and D, all gases) reacting in a closed flask according to the equation: $$\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g)$$ a. You add more A to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer. b. You have the original setup at equilibrium, and you add more D to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.

For the reaction $\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g),$ consider two possibilities: (a) you mix 0.5 mole of each reactant, allow the system to come to equilibrium, and then add another mole of \(\mathrm{H}_{2}\) and allow the system to reach equilibrium again, or \((b)\) you \(\operatorname{mix} 1.5\) moles of \(\mathrm{H}_{2}\) and 0.5 mole of \(\mathrm{I}_{2}\) and allow the system to reach equilibrium. Will the final equilibrium mixture be different for the two procedures? Explain.

For the reaction $$2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ it is determined that, at equilibrium at a particular temperature, the concentrations are as follows: $[\mathrm{NO}(g)]=8.1 \times 10^{-3} \mathrm{M}\( \)\left[\mathrm{H}_{2}(g)\right]=4.1 \times 10^{-5} M,\left[\mathrm{N}_{2}(g)\right]=5.3 \times 10^{-2} M,$ and \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]-2.9 \times 10^{-3} \mathrm{M} .\) Calculate the value of \(K\) for the reaction at this temperature.
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